Growth and Decay

Growth and Decay. Practice questions to deepen understanding of growth and decay. Online math practice with full solutions and step-by-step explanations.

Growth and decay practice — exponential functions, exponential growth, radioactive decay, half-life, real-life applications.

50 questions

Question 1
2.00 pts

📈 ?

?

Explanation:

💡 :

= ! 📈

<
Question 2
2.00 pts

📉 ?

?

Explanation:

💡 :

= ! 📉

Question 3
2.00 pts

📐 :

/?

Explanation:

💡 :

! 📐

f(t) = f(0) × qᵗ
Question 4
2.00 pts

🔍 ?

?

Explanation:

💡 :

-q! 🔍

📈
Question 5
2.00 pts

🎯 The initial quantity:

An experiment starts with 200 bacteria.

What is the value of f(0) in the formula?

Explanation:

💡 Detailed explanation:

f(0) = the starting point! 🎯

f(0) is the notation for the initial quantity

It is the number we have at time zero (t=0)

In our example:
"start with 200 bacteria"

So f(0) = 200

Verification:
f(0) = 200 × q⁰ = 200 × 1 = 200 ✓
Question 6
2.00 pts

⏰ The time variable:

What does t represent in the formula f(t) = f(0) × q^t?

Explanation:

💡 Detailed explanation:

t = number of time intervals! ⏰

t is the elapsed time

Important! t is measured in intervals

Examples:

• If the multiplier is "every hour":
then t = number of hours

• If the multiplier is "every year":
then t = number of years

• If the multiplier is "every 10 minutes":
then t = number of 10-minute units

Example:
Bacteria are multiplied ×2 every hour.
After 5 hours: t = 5
Question 7
2.00 pts

🔢 The multiplier (ratio):

The bacteria quantity is multiplied by 3 every hour.

What is the value of q?

Explanation:

💡 Detailed explanation:

q = the multiplier per interval! 🔢

q is the number we multiply by

"multiplied by 3" = multiply by 3

So q = 3

Examples:

📈 Growth:
• "multiplied by 2" → q = 2
• "increases by 50%" → q = 1.5
• "grows by factor 4" → q = 4

📉 Decay:
• "drops to half" → q = 0.5
• "80% remains" → q = 0.8
• "decays 25%" → q = 0.75
Question 8
2.00 pts

🔍 Identification:

Quantity at time 0: 100
Quantity at time 1: 150
Quantity at time 2: 225

What is happening here?

Explanation:

💡 Detailed explanation:

Finding q! 🔍

Data:
t=0: 100
t=1: 150
t=2: 225

Find q:
q = 150 ÷ 100 = 1.5

Verification:
225 ÷ 150 = 1.5 ✓

Conclusion:
• q = 1.5 > 1 → Growth! 📈
• Quantity grows by factor 1.5 (50%) each interval
Question 9
2.00 pts

🔍 Identification:

Drug quantity in body:
t=0: 400 mg
t=1: 320 mg
t=2: 256 mg

What is happening here?

Explanation:

💡 Detailed explanation:

Identifying decay! 🔍

Data:
t=0: 400
t=1: 320
t=2: 256

Find q:
q = 320 ÷ 400 = 0.8

Verification:
256 ÷ 320 = 0.8 ✓

Conclusion:
• q = 0.8 < 1 → Decay! 📉
• 80% of the quantity remains (drops 20%)
• Quantity decreases each interval
Question 10
2.00 pts

🔗 :

/ geometric sequence?

Explanation:

💡 :

! 🔗

Question 11
2.00 pts

📈 Find q:

Number of bacteria at t=0: 50
Number of bacteria at t=1: 100

What is q?

Explanation:

💡 Detailed explanation:

Finding q from data 📝

Given:
f(0) = 50 (t=0)
f(1) = 100 (t=1)

Compute q:
q = f(1) ÷ f(0)
q = 100 ÷ 50
q = 2

Meaning:
The quantity is multiplied ×2 each time interval

Verification:
f(1) = 50 × 2¹ = 100 ✓
Question 12
2.00 pts

📈 Computing growth:

Start with 200 bacteria.
The quantity is multiplied ×3 every hour.

How many bacteria are there after 2 hours?

Explanation:

💡 Detailed explanation:

Using the formula 📝

Data:
f(0) = 200
q = 3
t = 2 hours

Formula:
f(t) = f(0) × qᵗ

Substitution:
f(2) = 200 × 3²
f(2) = 200 × 9
f(2) = 1,800

Counting:
t=0: 200
t=1: 600
t=2: 1,800 ✓
Question 13
2.00 pts

📈 Growth of 20%:

A city has 10,000 residents.
The population grows by 20% each year.

How many residents will there be after one year?

Explanation:

💡 Detailed explanation:

Growth percentages! 📝

Understanding:
"growth of 20%" = 100% remains + 20% = 120%

Data:
f(0) = 10,000
Growth: 20% = 0.20
q = 1 + 0.20 = 1.2
t = 1

Computation:
f(1) = 10,000 × 1.2¹
f(1) = 10,000 × 1.2
f(1) = 12,000

⭐ Growth of 20% → q = 1.2
Question 14
2.00 pts

🔍 Find f(0):

After 3 hours there are 800 bacteria.
The multiplier is q = 2.

How many bacteria were there at the start?

Explanation:

💡 Detailed explanation:

Working backwards! 📝

Given:
f(3) = 800
q = 2
t = 3

Formula:
f(t) = f(0) × qᵗ

Substitution:
800 = f(0) × 2³
800 = f(0) × 8
f(0) = 800 ÷ 8
f(0) = 100

Verification:
100 → 200 → 400 → 800 ✓
Question 15
2.00 pts

⏰ After how long?

Start with 5 bacteria.
Every hour the quantity is multiplied ×2.

After how many hours will there be 80 bacteria?

Explanation:

💡 Detailed explanation:

Finding the time! ⏰

Data:
f(0) = 5
q = 2
f(t) = 80

Method 1: counting
t=0: 5
t=1: 10
t=2: 20
t=3: 40
t=4: 80 ✓

Method 2: formula
80 = 5 × 2ᵗ
16 = 2ᵗ
2⁴ = 2ᵗ
t = 4
Question 16
2.00 pts

📈 Computation:

f(0) = 100, q = 1.5

What is f(3)?

Explanation:

💡 Detailed explanation:

Direct computation 📝

Data:
f(0) = 100
q = 1.5
t = 3

Computation:
f(3) = 100 × 1.5³
f(3) = 100 × 3.375
f(3) = 337.5

Counting:
100 → 150 → 225 → 337.5 ✓
Question 17
2.00 pts

📈 Growth of 50%:

$200 in the account.
The money grows by 50% each year.

How much will there be after two years?

Explanation:

💡 Detailed explanation:

Growth of 50% 💰

Understanding:
Growth of 50%:
q = 1 + 0.50 = 1.5

Data:
f(0) = 200
q = 1.5
t = 2

Computation:
f(2) = 200 × 1.5²
f(2) = 200 × 2.25
f(2) = $450

Counting:
200 → 300 → 450 ✓
Question 18
2.00 pts

📊 Comparison:

Start with 10 bacteria, q = 3.

How many times more bacteria are there at t=4 compared to t=2?

Explanation:

💡 Detailed explanation:

Comparison! 📊

Computation:
f(2) = 10 × 3² = 10 × 9 = 90
f(4) = 10 × 3⁴ = 10 × 81 = 810

Ratio:
810 ÷ 90 = 9

Explanation:
Between t=2 and t=4, 2 intervals passed
So the quantity grew by q² = 3² = 9 times

⭐ Rule:
Multiplication by q each interval
Question 19
2.00 pts

📈 Computation:

Start with 3.
The quantity is multiplied ×4 each time.

What is the quantity after 3 intervals?

Explanation:

💡 Detailed explanation:

Solution 📝

Data:
f(0) = 3
q = 4
t = 3

Computation:
f(3) = 3 × 4³
f(3) = 3 × 64
f(3) = 192

Counting:
3 → 12 → 48 → 192 ✓
Question 20
2.00 pts

🔍 Find q:

At t=0: f=10
At t=3: f=80

What is q?

Explanation:

💡 Detailed explanation:

Finding q from 2 points! 🔍

Given:
f(0) = 10
f(3) = 80

Formula:
80 = 10 × q³
8 = q³
q = ∛8
q = 2

Verification:
10 × 2³ = 10 × 8 = 80 ✓

Counting:
10 → 20 → 40 → 80 ✓
Question 21
2.00 pts

📈 Growth of 25%:

A company has 80 employees.
The number of employees grows by 25% each year.

How many employees will there be after two years?

Explanation:

💡 Detailed explanation:

Solution 📝

Understanding:
Growth of 25%:
q = 1 + 0.25 = 1.25

Data:
f(0) = 80
q = 1.25
t = 2

Computation:
f(2) = 80 × 1.25²
f(2) = 80 × 1.5625
f(2) = 125

Counting:
80 → 100 → 125 ✓
Question 22
2.00 pts

🔍 How many at the start?

After 4 hours there are 810 bacteria.
q = 3.

How many were there at the start?

Explanation:

💡 Detailed explanation:

Working backwards 📝

Given:
f(4) = 810
q = 3

Solution:
810 = f(0) × 3⁴
810 = f(0) × 81
f(0) = 810 ÷ 81
f(0) = 10

Verification:
10 → 30 → 90 → 270 → 810 ✓
Question 23
2.00 pts

📈 Rapid growth:

A virus spreads on a computer.
Start with 2 infected files.
The quantity is multiplied ×5 every minute.

How many files are infected after 3 minutes?

Explanation:

💡 Detailed explanation:

Rapid growth! 🦠

Data:
f(0) = 2
q = 5
t = 3 minutes

Computation:
f(3) = 2 × 5³
f(3) = 2 × 125
f(3) = 250

Counting:
2 → 10 → 50 → 250 ✓

⚠️ Very rapid growth!
Question 24
2.00 pts

⏰ When does it reach?

Start with 20.
q = 2.

After how many intervals will there be 320?

Explanation:

💡 Detailed explanation:

Finding t 📝

Data:
f(0) = 20
q = 2
f(t) = 320

Counting:
20 → 40 → 80 → 160 → 320

After 4 intervals!

Or by formula:
320 = 20 × 2ᵗ
16 = 2ᵗ
2⁴ = 2ᵗ
t = 4
Question 25
2.00 pts

📈 Growth of 10%:

House price: $1,000,000.
The price rises by 10% each year.

What is the price after 3 years?

Explanation:

💡 Detailed explanation:

Growth of 10% 🏠

Understanding:
Growth of 10%:
q = 1 + 0.10 = 1.1

Data:
f(0) = 1,000,000
q = 1.1
t = 3

Computation:
f(3) = 1,000,000 × 1.1³
f(3) = 1,000,000 × 1.331
f(3) = $1,331,000
Question 26
2.00 pts

📊 :

:
A: 100 , 50%
B: 150 , 20%

?

Explanation:

💡 :

! 📊

A
Question 27
2.00 pts

🔢 Find q:

A quantity grows by 75% in each interval.

What is q?

Explanation:

💡 Detailed explanation:

Conversion from percent to q! 📝

Understanding:
"growth of 75%"

Meaning: the new quantity is:
100% (original) + 75% (added) = 175%

Compute q:
q = 1 + 0.75
q = 1.75

⭐ Rule:
Growth of x% → q = 1 + (x/100)
Question 28
2.00 pts

⏰ Doubling time:

A quantity is multiplied ×2 every 3 hours.
Start with 50.

How much will there be after 9 hours?

Explanation:

💡 Detailed explanation:

Important! The time interval! ⏰

Understanding:
The multiplier is "every 3 hours"
So t is measured in units of 3 hours

9 hours = how many intervals?
9 ÷ 3 = 3 intervals

Computation:
f(0) = 50
q = 2
t = 3

f(3) = 50 × 2³
f(3) = 50 × 8
f(3) = 400
Question 29
2.00 pts

📊 Comparison:

f(0) = 100, q = 1.2

What is the difference between f(5) and f(3)?

Explanation:

💡 Detailed explanation:

Computing the difference! 📝

Compute f(3):
f(3) = 100 × 1.2³
f(3) = 100 × 1.728
f(3) ≈ 173

Compute f(5):
f(5) = 100 × 1.2⁵
f(5) = 100 × 2.488
f(5) ≈ 249

Difference:
249 - 173 = 76
Question 30
2.00 pts

🌱 Problem:

A tree of height 2 meters.
Each year the height grows by 30%.

What will the height be after 4 years?
(round to whole number)

Explanation:

💡 Detailed explanation:

Tree growth! 🌱

Data:
f(0) = 2 meters
Growth: 30%
q = 1 + 0.30 = 1.3
t = 4 years

Computation:
f(4) = 2 × 1.3⁴
f(4) = 2 × 2.8561
f(4) ≈ 5.71
f(4) ≈ 6 meters
Question 31
2.00 pts

📉 Find q:

Drug quantity at t=0: 200 mg
Drug quantity at t=1: 100 mg

What is q?

Explanation:

💡 Detailed explanation:

Finding q in decay 📝

Given:
f(0) = 200 (t=0)
f(1) = 100 (t=1)

Compute q:
q = f(1) ÷ f(0)
q = 100 ÷ 200
q = 0.5

Meaning:
q < 1 → decay! 📉
50% of the quantity remains (drops 50%)

Verification:
f(1) = 200 × 0.5¹ = 100 ✓
Question 32
2.00 pts

📉 Decay computation:

There are 800 mg of drug in the body.
Each hour 0.5 of the quantity remains.

How much drug remains after 3 hours?

Explanation:

💡 Detailed explanation:

Using the formula 💊

Data:
f(0) = 800
q = 0.5
t = 3 hours

Formula:
f(t) = f(0) × qᵗ

Substitution:
f(3) = 800 × 0.5³
f(3) = 800 × 0.125
f(3) = 100 mg

Counting:
t=0: 800
t=1: 400
t=2: 200
t=3: 100 ✓
Question 33
2.00 pts

📉 Drop of 20%:

A car worth $100,000.
The value drops by 20% each year.

What is the value after one year?

Explanation:

💡 Detailed explanation:

Decay percentages! 🚗

Understanding:
"drops by 20%" = 100% - 20% = 80% remains

Data:
f(0) = 100,000
Drop: 20% = 0.20
q = 1 - 0.20 = 0.8
t = 1

Computation:
f(1) = 100,000 × 0.8¹
f(1) = 100,000 × 0.8
f(1) = $80,000

⭐ Drop of 20% → q = 0.8
Question 34
2.00 pts

🔍 Find f(0):

After 4 hours, 50 mg of drug remained.
q = 0.5.

How much drug was there at the start?

Explanation:

💡 Detailed explanation:

Working backwards! 📝

Given:
f(4) = 50
q = 0.5
t = 4

Formula:
f(t) = f(0) × qᵗ

Substitution:
50 = f(0) × 0.5⁴
50 = f(0) × 0.0625
f(0) = 50 ÷ 0.0625
f(0) = 800 mg

Verification:
800 → 400 → 200 → 100 → 50 ✓
Question 35
2.00 pts

⏰ After how long?

Start with 320 mg of drug.
q = 0.5 (half-life).

After how many hours will 40 mg remain?

Explanation:

💡 Detailed explanation:

Finding the time! ⏰

Data:
f(0) = 320
q = 0.5
f(t) = 40

Method 1: counting
t=0: 320
t=1: 160
t=2: 80
t=3: 40 ✓

Method 2: formula
40 = 320 × 0.5ᵗ
0.125 = 0.5ᵗ
0.5³ = 0.5ᵗ
t = 3
Question 36
2.00 pts

📉 Computation:

f(0) = 1000, q = 0.8

What is f(2)?

Explanation:

💡 Detailed explanation:

Direct computation 📝

Data:
f(0) = 1000
q = 0.8
t = 2

Computation:
f(2) = 1000 × 0.8²
f(2) = 1000 × 0.64
f(2) = 640

Counting:
1000 → 800 → 640 ✓
Question 37
2.00 pts

📉 Drop of 30%:

A car worth $200,000.
The value drops by 30% each year.

What is the value after two years?

Explanation:

💡 Detailed explanation:

Decay of 30% 🚗

Understanding:
Drop of 30%:
q = 1 - 0.30 = 0.7

Data:
f(0) = 200,000
q = 0.7
t = 2

Computation:
f(2) = 200,000 × 0.7²
f(2) = 200,000 × 0.49
f(2) = $98,000

Counting:
200,000 → 140,000 → 98,000 ✓
Question 38
2.00 pts

☢️ Half-life:

Radioactive material: 1000 grams.
Half-life = 5 years (every 5 years half remains).

How much remains after 15 years?

Explanation:

💡 Detailed explanation:

Half-life! ☢️

Understanding:
Half-life = the time in which half remains
q = 0.5

How many intervals?
15 years ÷ 5 years = 3 intervals

Computation:
f(0) = 1000
q = 0.5
t = 3

f(3) = 1000 × 0.5³
f(3) = 1000 × 0.125
f(3) = 125 grams
Question 39
2.00 pts

🔍 Find q:

At t=0: f=1000
At t=2: f=640

What is q?

Explanation:

💡 Detailed explanation:

Finding q from 2 points! 🔍

Given:
f(0) = 1000
f(2) = 640

Formula:
640 = 1000 × q²
0.64 = q²
q = √0.64
q = 0.8

Verification:
1000 × 0.8² = 1000 × 0.64 = 640 ✓

Counting:
1000 → 800 → 640 ✓
Question 40
2.00 pts

☕ Cooling:

Coffee at 80°C.
Each minute the temperature drops by 10%.

What is the temperature after 3 minutes?
(round to whole number)

Explanation:

💡 Detailed explanation:

Coffee cooling! ☕

Data:
f(0) = 80°C
Drop: 10%
q = 1 - 0.10 = 0.9
t = 3 minutes

Computation:
f(3) = 80 × 0.9³
f(3) = 80 × 0.729
f(3) = 58.32
f(3) ≈ 58°C
Question 41
2.00 pts

🔢 Find the decay percentage:

q = 0.85

By what percent does the quantity drop in each interval?

Explanation:

💡 Detailed explanation:

Conversion from q to percent! 📝

Given:
q = 0.85

Understanding:
q = 0.85 means 85% remains

So how much drops?
100% - 85% = 15%

⭐ Rule:
Decay percent = (1 - q) × 100%

Example:
q = 0.85 → decay of 15%
Question 42
2.00 pts

⏰ When does it reach?

Start with 2000 mg.
q = 0.5.

After how many intervals will 250 mg remain?

Explanation:

💡 Detailed explanation:

Finding t 📝

Data:
f(0) = 2000
q = 0.5
f(t) = 250

Counting:
2000 → 1000 → 500 → 250

After 3 intervals!

Or by formula:
250 = 2000 × 0.5ᵗ
0.125 = 0.5ᵗ
0.5³ = 0.5ᵗ
t = 3
Question 43
2.00 pts

📉 Drop of 15%:

A computer worth $5,000.
The value drops by 15% each year.

What is the value after 3 years?
(round to hundreds)

Explanation:

💡 Detailed explanation:

Decay of 15% 💻

Understanding:
Drop of 15%:
q = 1 - 0.15 = 0.85

Data:
f(0) = 5,000
q = 0.85
t = 3

Computation:
f(3) = 5,000 × 0.85³
f(3) = 5,000 × 0.614125
f(3) = 3,070.625
f(3) ≈ $3,100
Question 44
2.00 pts

📊 Comparison:

f(0) = 500, q = 0.8

How many times more is there at t=1 compared to t=3?

Explanation:

💡 Detailed explanation:

Comparison! 📊

Computation:
f(1) = 500 × 0.8¹ = 400
f(3) = 500 × 0.8³ = 500 × 0.512 = 256

Ratio:
400 ÷ 256 = 1.5625

Explanation:
Between t=1 and t=3, 2 intervals passed
In decay, the quantity drops by q² = 0.8² = 0.64
So the ratio is 1 ÷ 0.64 = 1.5625
Question 45
2.00 pts

🔍 How many at the start?

After 3 hours, 343 mg remained.
q = 0.7.

How many were there at the start?

Explanation:

💡 Detailed explanation:

Working backwards 📝

Given:
f(3) = 343
q = 0.7

Solution:
343 = f(0) × 0.7³
343 = f(0) × 0.343
f(0) = 343 ÷ 0.343
f(0) = 1000 mg

Verification:
1000 → 700 → 490 → 343 ✓
Question 46
2.00 pts

⏰ Time to half:

Start with 400.
q = 0.8.

After how many intervals will less than 200 remain?

Explanation:

💡 Detailed explanation:

Counting until we get there! ⏰

Counting:
t=0: 400
t=1: 320
t=2: 256
t=3: 204.8
t=4: 163.84 < 200 ✓

Answer:
After 4 intervals
The quantity will be less than 200

💡 At t=3 still above 200!
Question 47
2.00 pts

💊 Problem:

Your body breaks down a drug by 25% each hour (q=0.75).
You took 400 mg at 8:00.

How much drug will remain at 11:00?
(round to whole number)

Explanation:

💡 Detailed explanation:

Time computation! 💊

Time elapsed:
From 8:00 to 11:00 = 3 hours

Data:
f(0) = 400
q = 0.75
t = 3

Computation:
f(3) = 400 × 0.75³
f(3) = 400 × 0.421875
f(3) = 168.75
f(3) ≈ 169 mg
Question 48
2.00 pts

📊 :

:
A: 150,000 $, 10%
B: 180,000 $, 15%

equal ?

Explanation:

💡 :

! 📊

Question 49
2.00 pts

🔢 Find q:

A quantity drops by 35% in each interval.

What is q?

Explanation:

💡 Detailed explanation:

Conversion from percent to q! 📝

Understanding:
"drops by 35%"

Meaning: the new quantity is:
100% (original) - 35% (drop) = 65%

Compute q:
q = 1 - 0.35
q = 0.65

⭐ Rule:
Drop of x% → q = 1 - (x/100)
Question 50
2.00 pts

🌡️ Complex problem:

A hot drink at 90°C.
Every 5 minutes the temperature drops by 20%.

What is the temperature after 15 minutes?
(round to whole number)

Explanation:

💡 Detailed explanation:

Time and decay computation! 🌡️

Understanding:
Interval = 5 minutes
15 minutes = 15 ÷ 5 = 3 intervals

Data:
f(0) = 90°C
Drop: 20%
q = 1 - 0.20 = 0.8
t = 3

Computation:
f(3) = 90 × 0.8³
f(3) = 90 × 0.512
f(3) = 46.08
f(3) ≈ 46°C